ScalingStacks

0NAF

Lemma 3.11. Let W±0=W±0​(τ,o)W^{0}_{\pm}=W^{0}_{\pm}(\tau,o) and W±x=W±x​(τ,o)W^{x}_{\pm}=W^{x}_{\pm}(\tau,o) be pairs of corresponding webs in ⟦L±0⟧\left\llbracket L^{0}_{\pm}\right\rrbracket and ⟦L±x⟧\left\llbracket L^{x}_{\pm}\right\rrbracket respectively. Further, let s∈{p,t}2​ns\in\{p,t\}^{2n} be a palindrome in which tt appears 2​k2k times, and consider W±1=W±1​(τ,o,s)W^{1}_{\pm}=W^{1}_{\pm}(\tau,o,s) and W±x−1=W±x−1​(τ,o,s)W^{x-1}_{\pm}=W^{x-1}_{\pm}(\tau,o,s) in ⟦L±1⟧\left\llbracket L^{1}_{\pm}\right\rrbracket and ⟦L±x−1⟧\left\llbracket L^{x-1}_{\pm}\right\rrbracket respectively. Then

R​2−​(W−0,W−1)\displaystyle R2_{-}(W^{0}_{-},W^{1}_{-}) =(−1)k​R​2+​(W+0,W+1)\displaystyle=(-1)^{k}R2_{+}(W^{0}_{+},W^{1}_{+})
R​2+−1​(W+x−1,W+x)\displaystyle R2^{-1}_{+}(W^{x-1}_{+},W^{x}_{+}) =(−1)k​R​2−−1​(W−x−1,W−x).\displaystyle=(-1)^{k}R2^{-1}_{-}(W^{x-1}_{-},W^{x}_{-}).
0NAG

Proof. In a single Reidemeister II move, the identity resolution is always sent to the identity resolution via the identity. The maps involving the resolution with two thick edges are negatives of each other, when comparing the two types of Reidemeister II moves with fixed order of crossings as in (3.2). ∎

Original mathematics by the credited authors. Source collection and HTML conversion remain in progress.

Scott Morrison, Kevin Walker, Paul Wedrich

Original source: arXiv:1907.12194v5